Answer
$\dfrac{x^2}{1}+\dfrac{y^2}{4}=1$
Work Step by Step
The general form of the equation of an ellipse can be defined as:
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ when $a \lt b$
Here, we have Major axis length:'$2b=4 \implies b=2$
The eccentricity $e$ is given as: $e=\dfrac{c}{b}=\dfrac{\sqrt 3}{2}$
so, $\dfrac{c}{2}=\dfrac{\sqrt 3}{2} \implies c=\sqrt 3$
$c=\sqrt {a^2-2^2} \implies \sqrt 3=\sqrt {a^2-4} $
or, $a=1$
Thus, the general form of an ellipse as follows:
$\dfrac{x^2}{1}+\dfrac{y^2}{(2)^2}=1$
or, $\dfrac{x^2}{1}+\dfrac{y^2}{4}=1$
